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Modular Specht modules need not be simple, and form heads can vanish

Statement refuted

Every nonzero modular Specht module $S_{k}^{λ} \subseteq M_{k}^{λ}$ is simple, and its invariant-form quotient $D^{λ} = S_{k}^{λ} / R^{λ}$ is nonzero; in particular the Gram matrix and its quotient detect nonzero Specht modules in every characteristic.

Facts & Assumptions

Given: A prime $p$ and a splitting $p$ -modular system $(K, O, k)$ for $S_{n}$ (A splitting p-modular system for a finite group is a p-modular system whose fraction and residue fields split the needed group algebras), with the base-changed Specht module $S_{k}^{λ} \subseteq M_{k}^{λ}$ and its quotient $D^{λ} = S_{k}^{λ} / R^{λ}$ (Integral Specht lattice and base change, Modular Specht form and radical quotient). For the first witness $p = 3$ , $n = 3$ , $λ = (2, 1)$ and the two standard tableaux $t = \begin{matrix} 1 & 2 \\ 3 \end{matrix}$ , $u = \begin{matrix} 1 & 3 \\ 2 \end{matrix}$ ; for the second witness $p = 2$ , $n = 4$ , $λ = (2, 2)$ .

[F1]

The standard polytabloids form a basis of $S_{k}^{λ}$ over every field $k$ , and $\dim_{k} S_{k}^{λ}$ is the number of standard $λ$ -tableaux (Integral Specht lattice and base change, Tableaux and standard tableaux). In particular $\dim_{k} S_{k}^{(2, 1)} = 2$ with basis $e_{t}, e_{u}$ , and $\dim_{k} S_{k}^{(2, 2)} = 2$ with basis the polytabloids of the standard tableaux $12 / 34$ and $13 / 24$ .

[F2]

The $(2, 1)$ -tabloids are the three tabloids $v_{1}, v_{2}, v_{3}$ , where $v_{i}$ is the tabloid whose singleton second row is ${i}$ ; they form a $k$ -basis of $M_{k}^{(2, 1)}$ , and $σ$ acts by relabelling the entries, so $σ \cdot v_{i} = v_{σ (i)}$ (Young subgroups, tabloids, and permutation modules).

[F3]

For a tableau $t$ one has $κ_{t} = \sum_{γ \in C_{t}} sgn (γ) γ$ and $e_{t} = κ_{t} \cdot {t}$ ; here $C_{t} = {1, (13)}$ for $t = 12 / 3$ and $C_{u} = {1, (12)}$ for $u = 13 / 2$ , and $(13) \cdot t$ and $(12) \cdot u$ both have singleton second row ${1}$ (Column antisymmetrizers, polytabloids, and Specht modules). Consequently $e_{t} = v_{3} - v_{1}$ and $e_{u} = v_{2} - v_{1}$ in $M_{k}^{(2, 1)}$ , for every field $k$ .

[F4]

The integral tabloid form $β$ has the tabloids as an orthonormal basis; its scalar extension $β_{k}$ is symmetric and nondegenerate, and $R^{λ} = S_{k}^{λ} \cap (S_{k}^{λ})^{⊥}$ with $\dim_{k} D^{λ} = {rank}_{k} (G_{λ} m o d p)$ (Integral tabloid form and Specht Gram matrix, Modular Specht form and radical quotient).

[F5]

The example Specht Gram rank for shape (2,2) computes $G_{(2, 2)} = (\begin{array}{cc} 4 & 2 \\ 2 & 4 \end{array})$ and its reductions: ${rank}_{2} (G_{(2, 2)} m o d 2) = 0$ , while $\dim_{k} S_{k}^{(2, 2)} = 2$ in every characteristic.

[F6]

For a $p$ -modular system as above, $D^{λ} \neq 0$ if and only if $λ$ is $p$ -regular (Nonzero modular Specht quotient criterion).

[F7]

A $k [S_{n}]$ -module is simple when it is nonzero and has no proper nonzero submodule; a subspace closed under the action is a submodule (Simple module: a nonzero module with no proper nonzero submodule, Submodule of a module).

Counterexample

technique · direct

1.1givenF1F2F3F7algebra

Take $k$ of characteristic $3$ and $λ = (2, 1)$ , so $n = 3$ . By [F1] and [F3] the module $S_{k}^{(2, 1)}$ has $k$ -basis $e_{t} = v_{3} - v_{1}$ , $e_{u} = v_{2} - v_{1}$ , and $e_{t} + e_{u} = (v_{3} - v_{1}) + (v_{2} - v_{1}) = v_{1} + v_{2} + v_{3} - 3 v_{1} = v_{1} + v_{2} + v_{3} = : w$ because $3 v_{1} = 0$ in characteristic $3$ . In particular $w \in S_{k}^{(2, 1)}$ and $w \neq 0$ , since its coefficients at the basis vectors $v_{1}, v_{2}, v_{3}$ are all $1$ . For every $σ \in S_{3}$ one has $σ \cdot w = v_{σ (1)} + v_{σ (2)} + v_{σ (3)} = w$ by [F2], so the one-dimensional subspace $k w \subseteq S_{k}^{(2, 1)}$ is a submodule. It is proper because $\dim_{k} S_{k}^{(2, 1)} = 2$ by [F1]. Hence $S_{k}^{(2, 1)}$ has a proper nonzero submodule and is not simple by [F7]; the characteristic- $3$ witness is a nonzero two-dimensional modular Specht module with a one-dimensional trivial submodule.

1.2givenF4F5F6algebra

Take $k$ of characteristic $2$ and $λ = (2, 2)$ , so $n = 4$ . By [F5] the reduction of the integral Gram matrix $G_{(2, 2)}$ modulo $2$ is the zero matrix, of rank $0$ ; by the dimension formula of [F4] this gives $\dim_{k} D^{(2, 2)} = {rank}_{2} (G_{(2, 2)} m o d 2) = 0$ , so $D^{(2, 2)} = 0$ while $S_{k}^{(2, 2)} \neq 0$ of dimension $2$ by [F5]. Equivalently, $(2, 2)$ has the part $2$ occurring twice and is $2$ -singular, so [F6] also predicts $D^{(2, 2)} = 0$ . Thus the Gram quotient of a nonzero modular Specht module can vanish.

2.1givenF6step 1.1step 1.2∎

Step 1.1 exhibits a nonzero modular Specht module that is not simple, and step 1.2 exhibits a nonzero modular Specht module whose invariant-form quotient is zero; the two failures are independent, since the first occurs for a $p$ -regular label (where $D^{(2, 1)} \neq 0$ by [F6]) and the second for a $p$ -singular one. Hence the statement refuted fails in both clauses, and no field-independent appeal to the ordinary-case simplicity or to nonvanishing of the form quotient is available in prime characteristic.

Remarks

The quotient in the characteristic- $3$ witness. Quotienting $S_{k}^{(2, 1)}$ by $k w$ leaves a one-dimensional module; from $(12) \cdot e_{t} = v_{3} - v_{2}$ and $v_{3} - v_{2} + e_{t} = 2 v_{3} - v_{1} - v_{2} = 3 v_{3} - w \equiv 0$ one computes $(12) \cdot (e_{t} + k w) = - e_{t} + k w$ in characteristic $3$ , so the transposition acts by $- 1$ on the quotient and the quotient is the sign representation. With step 1.1 this is the factor list $[S_{k}^{(2, 1)}] = [D^{(3)}] + [D^{(2, 1)}]$ at $p = 3$ , matching the decomposition matrix of James Example 12.4 for $S_{3}$ (Dominance unitriangularity of the symmetric-group decomposition matrix).
What is not claimed about the characteristic- $2$ witness. Step 1.2 only refutes nonvanishing of the form quotient for $S_{k}^{(2, 2)}$ ; nothing here asserts that $S_{k}^{(2, 2)}$ is or is not simple in characteristic $2$ .
No detection criterion is claimed. The reduction of the Gram matrix computes $\dim_{k} D^{λ}$ and hence detects whether $D^{λ} \neq 0$ (Modular Specht form and radical quotient); it does not detect reducibility of $S_{k}^{λ}$ , as the characteristic- $3$ witness shows, where $D^{(2, 1)} \neq 0$ and yet $S_{k}^{(2, 1)}$ is reducible.

Depends on

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Sources

G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notes in Mathematics 682, Example 5.1 and Example 12.4, printed pp. 18 and 43 (standard reference, not scraped)
David A. Craven, Groups, Geometries and Representation Theory, §2.3 and Exercise 2.4, printed pp. 23-25 (standard reference, not scraped)